Icca Interview Questions, Process, and Tips

Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:

Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.

Now you have 4 litres of water in jug A.

Total game played= 60
%won =30%
Total won= 60*30/100 i.e. 18
now team plays x games and win all of those to increase the
average to 50%.
So,
(60+x)*50/100=18+x
(60+x)/2=18+x
60+x=36+2x
24=x
So the final answer is 24.

Ratio is 3:2:1

Total=6

Reverse of ratio is 1/3:1/2:1/1

Portion of first part is=1/3*6=>2

Portion of second part is=1/2*6=>3

Portion of third part is 1*6=>6

∴So the new ratio is 2:3:6

26 rupees and 53 paise

Tuesday

5436

6/11 or 54.54%

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins

let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9 so X<=3 and any of the digit in the number is <=9 and also given that 3 pairs sum is 11... so make trial and error.. if X=1...any of the no is 10 which is wrong trial.... if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296

Rs.265.80

The local value of 7 in the number is at 10000 so 7×10000 and face of 7 is 7
So 70000 -7 so the answer is C 69993